How do I integrate by substitution...?

Question

4X / [X^2 - 1]^1/2

So that is 4X over square root of (Xsquared) minus 1.

the substitution to use is 'u = x^2 -1'

Answer 1

INT( 4x/sqrt(x^2-1) dx

INT 2(2x dx/(sqrt(x^2-1)

put x^2-1 = u

2x dx = du

INT 2(du/sqrt(u)

2(sqrt(u)/(1/2) + c

4sqrt(u) + c

4sqrt(x^2-1) + c

Answer 2

this is a fairly easy integral compared to others that you probably will encounter in calc II.

we have ∫4x/(x^2-1)^(1/2), notice that the 4 in front of the x is a constant, so it could be taken out and put in front of the integral sign. 4∫x/(x^2-1)^(1/2), now let u = x^2-1 and the derivative of u du = 2x dx, and 1/2du = xdx. Now the problem becomes 4∫(1/2du)/u^(1/2) notice again that the 1/2 is a constant that you could take out and put in front of the integral, so this gives you 1/2*4∫du/u^(1/2) = 2∫du/u^(1/2) = 2∫u^(-1/2)du and now add one to the exponent on u, and devide by the new exponent, this gives us 2[2u^(1/2)] +c = 4u^(1/2) +c and finally substitute back in for u, and the answer is:

4(x^2-1)^1/2 +c = 4sqrt(x^2-1) + c

the c at the and is just some constant that we always add after we do an indefinite integral.

Answer 3

Well that's it, if you must use substitution.
If u = x² - 1 then du = 2xdx
So the integral becomes
∫(2/u^½)du = 2∫(u^-½)du = 4u^½ + c
=4√(x² - 1) + c

Easier to use the "function of a function" rule (aka chain rule) to differentiate (x² - 1)^½ , which gives x(x² - 1)^-½ So the integral must be 4(x² - 1)^½ + c, since integration is the reverse of differentiation.

Answer 4

u = x ² - 1
du = 2 x dx
2 du = 4x dx
I = ∫ 4x / (x² - 1)^(1/2) dx
I = 2 ∫ 1 / u^(1/2) du
I = 2 ∫ u^( - 1/2 ) du
I = 4 u^(1/2) + C
I = 4 ( x ² - 1 )^(1/2) + C